<p>首先,感谢约翰·德瑞科先生为interparc服务。多好的工作啊!在</p>
<p>我也面临这个问题,但不熟悉MATLAB引擎API。鉴于此,我尝试将部分interparc Matlab代码转换为Python(只包括线性插入,因为它足以解决我的问题)。在</p>
<p>下面是我的代码;希望它能帮助所有寻求类似东西的Python:</p>
<pre><code>import numpy as np
def interpcurve(N,pX,pY):
#equally spaced in arclength
N=np.transpose(np.linspace(0,1,N))
#how many points will be uniformly interpolated?
nt=N.size
#number of points on the curve
n=pX.size
pxy=np.array((pX,pY)).T
p1=pxy[0,:]
pend=pxy[-1,:]
last_segment= np.linalg.norm(np.subtract(p1,pend))
epsilon= 10*np.finfo(float).eps
#IF the two end points are not close enough lets close the curve
if last_segment > epsilon*np.linalg.norm(np.amax(abs(pxy),axis=0)):
pxy=np.vstack((pxy,p1))
nt = nt + 1
else:
print('Contour already closed')
pt=np.zeros((nt,2))
#Compute the chordal arclength of each segment.
chordlen = (np.sum(np.diff(pxy,axis=0)**2,axis=1))**(1/2)
#Normalize the arclengths to a unit total
chordlen = chordlen/np.sum(chordlen)
#cumulative arclength
cumarc = np.append(0,np.cumsum(chordlen))
tbins= np.digitize(N,cumarc) # bin index in which each N is in
#catch any problems at the ends
tbins[np.where(tbins<=0 | (N<=0))]=1
tbins[np.where(tbins >= n | (N >= 1))] = n - 1
s = np.divide((N - cumarc[tbins]),chordlen[tbins-1])
pt = pxy[tbins,:] + np.multiply((pxy[tbins,:] - pxy[tbins-1,:]),(np.vstack([s]*2)).T)
return pt
</code></pre>